The centres of a set of circles, each of radius 3, lie on the circle `x^2+y^2+25`. The locus of any point in the set is: (a) `4lex^(2)+y^(2)le64` (b) `x^(2)+y^(2)le25` (c) `x^(2)+y^(2)ge25` (d) `3 lex^(2)+y^(2)le9`

Options

A. `4lex^(2)+y^(2)le64`
B. `x^(2)+y^(2)le25`
C. `x^(2)+y^(2)ge25`
D. `3 lex^(2)+y^(2)le9`

Explanation

To solve the problem, we need to find the locus of any point in a set of circles whose centers lie on the circle defined by the equation \(x^2 + y^2 = 25\) and each having a radius of 3.

### Step-by-Step Solution:

1. **Identify the Circle of Centers**:
The centers of the circles lie on the circle given by the equation:
\[ x^2 + y^2 = 25 \]
This means that the centers of the circles are at a distance of 5 from the origin (0,0).

2. **Determine the Radius of the Circles**:
Each circle has a radius of 3.

3. **Calculate the Minimum and Maximum Distances from the Origin**:
- The **minimum distance** from the origin to the edge of any circle is:
\[ 5 - 3 = 2 \]
- The **maximum distance** from the origin to the edge of any circle is:
\[ 5 + 3 = 8 \]

4. **Formulate the Inequality for the Locus**:
The locus of points that can be reached by the edge of the circles is therefore bounded by the distances calculated:
\[ 2 \leq \sqrt{x^2 + y^2} \leq 8 \]
Squaring all parts of the inequality gives:
\[ 4 \leq x^2 + y^2 \leq 64 \]

5. **Final Locus Equation**:
This can be expressed as:
\[ 4 \leq x^2 + y^2 \leq 64 \]
This means the locus of any point in the set is the area between the circles of radius 2 and radius 8 centered at the origin.

6. **Select the Correct Option**:
Looking at the provided options, the correct one that matches our derived inequality is:
\[ 4 \leq x^2 + y^2 \leq 64 \]
Therefore, the answer is:
**(a) \(4 \leq x^2 + y^2 \leq 64\)**.